Find $x$ as function of $y$

$\begingroup$

I am trying to find $y$ as function of $x$ of: $$y=\frac{x-1}{x+1}$$

I have tried $$y=\frac{x-1}{x+1}\iff (x+1)y=x-1\iff(x+1)y+1=x$$

Which is of course wrong

$\endgroup$ 3

2 Answers

$\begingroup$

You are almost there:

$$(x+1)y+1=x$$ $$xy + y + 1 = x$$ $$(xy -x) + (y + 1) = 0$$ $$(y-1)x+(y+1)=0$$ $$(y-1)x= -(y+1)$$ $$x=-\frac{y+1}{y-1}$$

EDIT per comment

Notice that from $y=\frac{x-1}{x+1}$, it is already guaranteed that $y \ne 1$, and thus we do not need to explicitly specify that.

$\endgroup$ 2 $\begingroup$

Multiplying by $1+x$ gives: \begin{align}yx+y&=x-1\\ yx-x&=-y-1\\ x(y-1)&=-1-y\end{align}

Therefore $$x=\frac{-1-y}{y-1}$$ if $y\ne 1$

$\endgroup$ 1